Question

According to a poll taken last year, 45% of the cities' residents get most of their news from the Internet. To conduct a follow-up study that would provide 90% confidence that the point estimate is correct to within 0.04 of the population proportion, how large a sample size is required

Answer 1

The sample size is
`n =419`

Explanation:

From the question we are told that

The population proportion is
`p = 0.45`

The margin of error is
`E = 0.04`

Given that the confidence level is 90%

Then the level of significance can be mathematically represented as

`\alpha = 100 -90`

`\alpha = 10\%`

`\alpha = 0.10`

Next we obtain the level of significance from the normal distribution table the value is

`Z_{(\alpha )/(2) } = 1.645`

Generally the sample size is mathematically represented as

`n = [ \frac{Z_{(\alpha )/(2) }}{E} ]^2 * p(1- p )`

substituting values

`n = [ (1.645 )/(0.04) ]^2 * 0.45(1- 0.45 )`

`n =419`